# Tag Info

Due to the amount of momentum (and energy) transferred to you. Assuming a completely inelastic process (and an idealized one dimensional setup) you get from the conservation of momentum ($v$ is the initial velocty, $p$ the initial momentum, $v'$ is the final velocity, $M$ is the mass of the object hitting you, $m$ is your mass): $$p = M v = (M + m) v'$$ $$... 1 The usual approach: write down conservation of angular momentum, linear momentum, and energy. Assume the impact is elastic and infinitely short duration. In that time the spring didn't move and the third particle didn't come into the equation. That means the problem can be reduced to two simpler problems: two particles that hit elastically (after collision ... 0 Completing the very good answer made by Andre Holzner, the electron and the positron in the initial state, have a non negligible probability to emit a photon with a significant energy. In particle physics jargon, this is called the ISR standing for Initial State Radiation. Therefore, you always have to be slightly above the threshold production to circumvent ... 3 In principle, you don't need to tune to 'exactly' this energy but having less than m_h + m_Z suppresses this diagram. Having more should typically give you a higher cross section because there is 'more phase space' the final state particles can be in, i.e. the 'excess' energy will just be used as kinetic energy of the final state particles (Z and h ... 0 Mainly because of surface tension, but ofcourse also because of Newtons laws. Water has a Surface tension of 72.8 mN/m, The unit, N/m can also be written kg/s^2 or J/m2 In this case the J/m2 is the most practical. The Smaller the hitting angle of the stone makes this area bigger providing more Energy to be returned on elastic collision. There is also ... 0 I think what you would do here is find the momentum. Momentum = Mass x Velocity. so lets say the objects speed is 10 mph, and the weight is 15 lbs. When the object in motion hits the resting object, it will deliver 150 lbs of force. If the object in motion is going 30 mph and weighs 2000 lbs (lets say a car) then the force it will deliver upon impact with ... 2 In a collision it's often the case that it's hard to measure exactly how long the collision lasts and exactly how the force between the objects changes during the collision. Squishy objects like nerf balls will collide relatively slowly while hard objects like billard balls will have a short collision time. However there is a well defined quantity called ... 4 You are not the first person to ask this question. http://superuser.com/questions/925826/what-would-put-a-hdd-under-350gs-of-force claims that 350 g of force is slightly more than a soccer player kicking a football. What this means is that you basically can kick your case, and it shouldn't brick your hard drive. It might cause other issues though, so don't ... 1 An elastic collision is defined as one which conserves energy. When you jump against a wall, most of your kinetic energy is dissipated as heat into your tissue as your legs and muscles absorb the impact. Therefore, energy is not conserved so by definition this is an inelastic collision. -2 A g is a unit of acceleration. I am going to assume the force meant here is 350 times the earth weight of the hard drive. Assuming the drive is 625 grams, that works out to a force of about 450 lbs. This is easy to achieve: just hit it with a hammer. Edit: as many have pointed out, the rating is given in units of acceleration for good reason. An ... 16 Here's an application where an ability to withstand high shock is important. Explosions. In the mid 1980s I did work for a mining company's research laboratory (BHP Research, now defunct like all Australian corporate research). We would lower data-logging computers into boreholes to set up a grid of dataloggers, then detonate a charge of known energy at a ... 44 Is it even possible to hit 350Gs of force to a hard drive? Sure is. Drop it on the floor. You are thinking about sustained forces. 350g sustained won't happen even in rocket launches. But momentary forces can easily peak at this level. Note that the G limit on the drive is for when it's not running. No spinning drive will like 350g, except maybe in ... 0 You need the integration step to be much shorter than the impact time. For most real world collisions, 0.1 second will be much too long. The force will change during the collision - and given the very simplistic integration method you use, you have to integrate over sufficiently short steps during which the force doesn't change much. You could do this by ... 1 The law states this: Two objects masses and with velocities m_1, m_2, \vec{v}_1 and \vec{v}_2 collide with contact normal \vec{n}. The final velocities are \vec{v}_1^\star and \vec{v}_2^\star such that the coefficient of restitution \epsilon is defined by$$\vec{n}\cdot \left( \vec{v}_2^\star - \vec{v}_1^\star \right) = -\epsilon \;\left( ...